A diatomic gas is used in car not heat engine having efficiency 80% . Find the ratio of initial volume to final volume of gas during adiabatic expansion.

Options

A. `((1)/( 5))^((3)/(2))`
B. `((1)/(3))^((5)/(2))`
C. `((1)/(5))^((5)/(2))`
D. `((1)/(5))^((2)/(5))`

Answer

To solve the problem of finding the ratio of the initial volume to the final volume of a diatomic gas during adiabatic expansion in a Carnot heat engine with an efficiency of 80%, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Efficiency of the Carnot Engine:**
The efficiency (\( \eta \)) of a Carnot engine is given by the formula:
\[ \eta = 1 - \frac{T_2}{T_1} \]
Given that the efficiency is 80%, we can write:
\[ 0.8 = 1 - \frac{T_2}{T_1} \]

2. **Rearranging the Efficiency Equation:**
Rearranging the equation gives:
\[ \frac{T_2}{T_1} = 1 - 0.8 = 0.2 \]
This implies:
\[ T_1 = 5T_2 \]

3. **Using the Adiabatic Process Relationship:**
For an adiabatic process, we have:
\[ T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]
where \( \gamma \) is the heat capacity ratio. For a diatomic gas, \( \gamma = \frac{C_p}{C_v} = \frac{7}{5} \).

4. **Substituting the Temperature Relationship:**
Substituting \( T_1 = 5T_2 \) into the adiabatic equation gives:
\[ 5T_2 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \]

5. **Canceling \( T_2 \):**
Since \( T_2 \) is common in both terms, we can cancel it out (assuming \( T_2 \neq 0 \)):
\[ 5 V_1^{\gamma - 1} = V_2^{\gamma - 1} \]

6. **Expressing the Volume Ratio:**
Rearranging the equation gives:
\[ \frac{V_1^{\gamma - 1}}{V_2^{\gamma - 1}} = \frac{1}{5} \]
Taking the \( \gamma - 1 \) root on both sides gives:
\[ \left(\frac{V_1}{V_2}\right)^{\gamma - 1} = \frac{1}{5} \]

7. **Substituting \( \gamma \):**
Substituting \( \gamma = \frac{7}{5} \) into the equation:
\[ \gamma - 1 = \frac{7}{5} - 1 = \frac{2}{5} \]
Thus:
\[ \left(\frac{V_1}{V_2}\right)^{\frac{2}{5}} = \frac{1}{5} \]

8. **Finding the Volume Ratio:**
Raising both sides to the power of \( \frac{5}{2} \):
\[ \frac{V_1}{V_2} = \left(\frac{1}{5}\right)^{\frac{5}{2}} = \frac{1}{\sqrt{125}} = \frac{1}{11.18} \approx 0.0894 \]

9. **Final Result:**
Therefore, the ratio of the initial volume to the final volume is:
\[ \frac{V_1}{V_2} = \frac{2}{5} \]

### Conclusion:
The ratio of the initial volume to the final volume of the gas during adiabatic expansion is \( \frac{2}{5} \).